Infinite square well expectation value problem

Fakestreet123
Messages
2
Reaction score
0

Homework Statement


A particle in an infinite box is in the first excited state (n=2). Obtain the expectation value 1/2<xp+px>


2. The attempt at a solution

Honestly, I don't even know where to begin.
I assumed V<0, V>L is V=∞ and 0<V<L is V=0

I tried setting up the expectation value formula

1/2∫x|ψ*Pψ|2+(x|ψ2)*P x|ψ|2

but what is ψ? is it √(2/L) Sin ((n2πx)/L) because its the solution to the infinite well? are the bounds from 0 to L?

Thanks for viewing my question!
 
Last edited:
Physics news on Phys.org
\psi is the wave function for the given problem, that you find by solving Schrodinger equation for a given potential
 
Fakestreet123 said:

Homework Statement


A particle in an infinite box is in the first excited state (n=2). Obtain the expectation value 1/2<xp+px>


2. The attempt at a solution

Honestly, I don't even know where to begin.
I assumed V<0, V>L is V=∞ and 0<V<L is V=0
You mean V=∞ when x<0 or x>L.

I tried setting up the expectation value formula

1/2∫x|ψ*Pψ|2+(x|ψ2)*P x|ψ|2
How did you get this?

but what is ψ? is it √(2/L) Sin ((n2πx)/L) because its the solution to the infinite well? are the bounds from 0 to L?
You do use a solution for the infinite square well, but the function you wrote isn't quite correct.
 
Thanks for responding guys! I very much appreciate it!

vela said:
You mean V=∞ when x<0 or
x
>L.

Ahh yes, sorry for the crappy notation


How did you get this?

Uhhh Now that I've done some more studying I can see that its wrong but I was, at the time, hoping that the expectation values would be ∫<xp>+<px> = ∫<xp+px> and the expectation values of xp, px will just add together like magic.

Which is wrong! So I've redone the problem in an attempt to simply find the operator first

<xp> = -ihx(dψ/dx) and
<px> = -ih (d(xψ)/dx)
1/2<xp+px> = (-ih)/2 (x(dψ/dx)+(d(xψ)/dx)

Now the expectation value should be <1/2(xp+px)>= ∫ψ*(1/2)<xp+px>ψ dx?

do I just plug in 1/2<xp+px> (assuming its correct) and plug in ψ (the solution to the infinite square well) and take the integral from 0 to L?

You do use a solution for the infinite square well, but the function you wrote isn't quite correct.

Sorry! I was very wrong haha
√(2/L) Sin ((nπx)/L)
 
One thing that might help you:

Using [x,p] = ihbar, or xp - px = ihbar, or px = -ihbar + xp, you can write

xp + px = xp + xp - ihbar = 2xp - ihbar, which has an easier to calculate expectation value. (you should check my work, I may have made a mistake)

In general an expectation value is:

<Q> = ∫ψ*Qψdx, where Q on the right is the operator for the observable you want to calculate. ψ is the wavefunction of the particle, which should be supplied by your textbook (it looks like you have the right one though).

Removed by moderator[/color]

Where I used the normalization condition ∫ψ*ψdx = 1.

I'll leave it to you do actually do the integral, and to check my work.
 
Last edited by a moderator:
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top